Question

A solenoid used to produce magnetic fields for research purposes is 2.2 m long, with an inner radius of 30 cm and 1300 turns of wire. When running, the solenoid produced a field of 1.2 T in the center. Part A Given this, how large a current does it carry? Express your answer in kiloamperes. I I = nothing kA SubmitRequest Answer Provide Feedback Next

Answer 1

Answer:

I= 1.616 KA

Explanation:

Given that

L = 2.2 m

N= 1300 turns

r= 30 cm

B= 1.2 T

We know that magnetic field in solenoid given as

$B=\dfrac{\mu _oIN}{L}$

$I=\dfrac{BL}{\mu_o N}$

Now by putting  the values

$I=\dfrac{1.2\times 2.2}{4\pi \times 10^{-7} \times 1300}\ A$

I= 1616.03 A

So the current is 1616.03 A.

I= 1.616 KA

Answer 2

The current passing the solenoid is 1.4kA

Data;

• l = 2.2m
• B = 1.2T
• μ = 4π*10^-7 TA^-1 m
• N = 1500
• r = 30cm = 0.3m

Magnetic Field Charge

Using the formula of magnetic field charge of a solenoid

$B=\frac{\mu_o NI}{L}\\I = \frac{BL}{\mu_o N}\\I= \frac{1.2*2.2}{4\pi *10^-^7*1500}\\I = 1400.56A\\I = 1.4kA$

From the calculations above, the current passing the solenoid is 1.4kA

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